The Thevenin theorem states that any real
source may be represented as an ideal potential source in series
with a resistor. In many cases, one may use the Thevenin circuit
to solve electronics problems that might otherwise be tedious at
best. Take, for example a "bridge" circuit shown in
Figure 1, wherein an element is placed such as to bridge two
parallel arms of a resistor network. We may need to know the
current through and potential across the bridging resistor, R5.

Figure 1 Resistor bridge circuit
with 25 volt source potential.

We start by removing (in theory) R5
to determine the Thevenin equivalent to the resistor network
composed of R1 through R4.
The resulting circuit may be used to determine the potential and
current through R5 upon inserting this element
into the Thevenin equivalent circuit. To make the example more
concrete, consider the following for resistance; R1=
1.0 kW
, R2=900 W , R3=800 W, and R4=1.1
kW .
The value of R5 does not have to specified to
derive a formula.

Figure 2 Bridge circuit with R5
removed.

The potential is more easily calculated since
the equivalent circuit reduces to two parallel voltage dividers.
Using Ohms law, one find the current through the series
arms, then the potential drop across the second resistor. For
example, the potential at point A is the current through
this arm times the resistance of R3. Thus

The Thevenin equivalent resistance found by
shorting-out the ideal potential source (in theory) and
determining resistance between points A and B.

Figure 3 Equivalent circuit with
voltage source short-circuited.

In this case the resistance is the sum of the
two parallel circuits

The resistance of the Thevenin equivalent
circuit can be found to be 939 W .

The resulting Thevenin equivalent circuit shown
in Figure 4. This circuit is functionally equivalent to that in
Figure 2.

Figure 4 The Thevenin equivalent
circuit of the bridge circuit with R5 restored.

It is now relatively easy to calculate the
current through the circuit. The total resistance for the series
resistors is RTOT=REQ+R5.
Thus the current through both the equivalent resistor and R5
is

I=EEQ /(REQ+R5).

The potential between points A and B
changes when R5 is restored. This new potential
is calculated using the voltage divider formula, or can be found
by considering the voltage drop across R5 due
to Ohms law. In either case